What is the Conservation of Energy in Spring Work at an Angle?

[hayke101]

Homework Statement

I'm trying to solve this question: A body is released from position (b), what will his speed at position (a)

https://imgur.com/a/Ms7widx

I know i need to use the spring energy formula but not sure how to do it at an angle. Will i have to multiply it by the angle of the spring. Or calculate the difference in length between the two positions?

Thanks :)

Relevant Equations

Work equation

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[Steve4Physics]

I know i need to use the spring energy formula but not sure how to do it at an angle. Will i have to multiply it by the angle of the spring. Or calculate the difference in length between the two positions

Hi @hayke101 and welcome to PF.

The rules here require you to show your attempt/thinking first. What are your thoughts?

 

[hayke101]

Hi @hayke101 and welcome to PF.

The rules here require you to show your attempt/thinking first. What are your thoughts?

I had quite a few attemps my current one is:
The energy at A is equal to the energy at B

At B with have no kenetic energy, and 2 spring energies
at A with have kenetic energy (and no spring energy?)
So 2*0.5kx^2 = 0.5mv^2

But I'm not sure if it is correct use x as the distance between A and B.

Also the spring is at an angle so maybe I need to add the angle someway (cos Ø?)

 

[PeroK]

But I'm not sure if it is correct use x as the distance between A and B.

I'm sure that definitely must be wrong! Neither spring is contracting by ##x##.

 

[hayke101]

I'm sure that definitely must be wrong! Neither spring is contracting by ##x##.

So they are contracting by ##s - l(0)##
Where s is the length of the spring at position B?

 

[PeroK]

So they are contracting by ##s - l(0)##
Where s is the length of the spring at position B?

That's definitely true. Can you simply use conservation of energy? Or, does the opposition of the two forces lose energy?

What about analysing the forces? Is that another way to solve the problem?

 

[Steve4Physics]

Can I add to what @PeroK has said. You are using the same symbol 'x' for two different things.

In your expression “0.5kx^2”, the 'x' means the spring’s extension (its increase in length from its unstretched length). But in your diagram, 'x' means the horizontal distance the mass moves.

Use a different symbol for the spring's extension (e.g. ‘e’). Edit1: Or use 'd'!
Edit2: no don't use 'd', as the question already uses 'd' for the initial extension.

 

[hayke101]

That's definitely true. Can you simply use conservation of energy? Or, does the opposition of the two forces lose energy?

What about analysing the forces? Is that another way to solve the problem?

I think that the opposition of forces doesn't "lose" energy and still keeps the conservation of energy.

I thought about analysing forces but the problem I came across is that i will need to do inegral on an equation with theta (which is kinda difficult I think)

Edit: btw thanks a lot for the help

 

[Steve4Physics]

View attachment 293566

Did you sort it out? If not (as I suspect from your attachment) try answering these...

Q1. The initial extension of each spring is d, so the initial energy stored in each spring is ____.

Q2. When m is displaced a distance x to the right, the new length of each spring, in terms of x and θ, is _______ (hint: use a bit of simple trigonometry).

Q3. Using the answer to Q2, the new extension of each spring is ___________.

Q4. Using the answer to Q3, the new energy stored in each spring is now _____________

Q5. So when m was displaced a distance x to thee right (and using your answers to Q1 and Q4) each spring was given additional energy _____________.

If you can answer those, the rest of the problem should be plain sailing.

 

[PeroK]

If you assume conservation of energy, what do you get?